\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow2x^2-11x+5-2x^2+10x=25\Leftrightarrow-x=20\Leftrightarrow x=-20\)

\(\Leftrightarrow\left(x-2021\right)\left(x-5\right)-\left(x-2021\right)=0\\ \Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\)

a: Ta có: \(\left(x-\dfrac{2}{5}\right)\left(x+\dfrac{2}{7}\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{2}{5}\\x< -\dfrac{2}{7}\end{matrix}\right.\)

⇔5-x=0;6+6x=0;2x-4=0

TH₁: 5-x=0           TH₂:6+6x=0              TH₃:2x-4=0

   ⇔x=5                ⇔x=-1                        ⇔x=2

       Vậy x∈{5;-1;2}

Đặt x^2=t

pt có 4 no pb=>pt2t^2-(m-1)t+m-3=0 có 2 no pb >0

=>\(\left\{{}\begin{matrix}\Delta>0\\P>0\\S>0\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}m^2-2m+1-4m+12>0\\\dfrac{m-3}{2}>0\\m-1>0\end{matrix}\right.\)=>...=>m>3

Vậy m>3

Δ=(-2)^2-4m=4-4m

Để phương trình có 2 nghiệm thì 4-4m>=0

=>4m<=4

=>m<=1

x1-x2=3

=>(x1-x2)^2=9

=>(x1+x2)^2-4x1x2=9

=>2^2-4m=9

=>4m=-5

=>m=-5/4(nhận)

`(4x+2)^2+(1-5x)^2-4(2x+1)(1-5x)=0`

`=> (4x+2)^2-4(2x+1)(1-5x)+(1-5x)^2=0`

`=> (4x+2-1+5x)^2=0`

`=> (9x+1)^2=0`

`=> 9x+1=0`

`=> 9x=-1`

`=> x= -1/9`

Vậy \(S=\left\{-\dfrac{1}{9}\right\}\)